Chapter 9.4, Problem 1QR

To determine

To find: the value of integral.

Answer to Problem 1QR

The value of integral is $\ln 2$ .

Explanation of Solution

Given information:

Theintegral is ${\displaystyle {\int }_0^3\frac{1}{x+3}dx}$ .

Calculation: to evaluate the integral, first integrate than substitute the upper and lower limit and simplify,

  $\begin{array}{l}{\displaystyle {\int }_0^3\frac{1}{x+3}dx}={\left[\ln \left(x+3\right)\right]}_0^3\text{ [integrate]}\\ {\displaystyle {\int }_0^3\frac{1}{x+3}dx}=\ln \left(3+3\right)-\ln \left(0+3\right)\text{ [substitute upper and lower limit]}\\ {\displaystyle {\int }_0^3\frac{1}{x+3}dx}=\ln 6-\ln 3\text{ [add]}\\ {\displaystyle {\int }_0^3\frac{1}{x+3}dx}=\ln 2+\ln 3-\ln 3\text{ [simplify]}\\ {\displaystyle {\int }_0^3\frac{1}{x+3}dx}=\ln 2\end{array}$

Thus, the value is $\ln 2$ .